liiWIiiilil 



T 






Class ___XJiJ_ 

Book . H '^'b 

Copyright ]^°_ 

COEKRIGHT DEPOSm 



Constructive Drawing. 

A TEXT-BOOK FOR HOME INSTRUCTION, HIGH SCHOOLS, MANUAL TRAINING 
SCHOOLS, TECHNICAL SCHOOLS AND UNIVERSITIES. 

AEEANGED BY 

HERMAN HANSTEIN, 

Supervisor of Drawing Chicago High Schools. Director of Art and Technical Drawing Departments Chicago Mechanics' 
Institute and Columbian Trade and Business School. 




KEUFFEL & 
ESSER CO. 



SECOND EDITION. ' ' ' 

GEOMETRIC CONSTRUCTIONS. 



CHICAGO 1904. 



NEW YORK 

CHICAGO 

ST. LOUIS 

SAN FRANCISCO 



LIBR^BV nl CONGRESS 

Two OoDies Recelrad 

SEP 3 1904 

, Cooyrlffht Entry 

ij^t, Id, / Tf 4- 

CLASS — KXe.No. 

COPY B 



T353 



ENTERED ACCORDING TO ACT OF CONGRESS, IN THE YEAR 1S9wt, 



3Y HERMAN HANS! 



IN THE OFFICE OF THE LIBRARIAN OF CONGRESS, AT WASHINGTON. 



PREFACE 



A T the request of assistants and pupils, as a help and for home instruction, I 

have compiled this course of constructive drawing, as it has been taught 

for the past twenty years in the Chicago City High-Schools, in the Drawing 

Department of the Chicago iviechanics' Institute and lately in the Columbian 

Trade and Business School. 

A practical experience of seventeen years in office and shop and his occupation 
as teacher during the past twenty years have given the author such experience 
and judgment as to select only such problems as are of practical importance to 
those who follow architectural, mechanical and engineering vocations as well as 
problems which are indispensable to manufacturing and industrial pursuits. 

As draughtsman, I have endeavored to arrange this work so as to bring into 
immediate application all the tools which are required in every drafting-room, 
and time and expense have not been spared to impress the student with the 
cardinal virtues of a successful draughtsman: accuracy and cleanliness. 

The author will feel well rewarded for his trouble if men of ability and 
learning will think it worth their while to point out to him any deficiencies that 
they may notice in perusing this work. HERMAN HAN STEIN 

Chicago, 111., September, 1894- 



Preface to Second Edition 



' I "HIS editition has been revised and for the convenierjce of both student and 
teacher the number of constructions on each plate has been reduced 
one half. 

In thanking colleagues for the kindness with which the first edition was 
received, the author hopes that in this new arrangement the "Geometric Con- 
structions" will retain the favor of those who have followed this course in the 
preparation of Projection and Descriptive Geometry. 

HERMAN HANSTEIN. 
Chicago, 111., June, I901f. 



INDEX. 



Tools, implements and their applications Figs. 1 to 7, Plate 1 

Alphabets Plate 2 

Block Alphabets Plate 3 

Bet squares and their applications Figs. 1 to 4, Plate 3 

Comers for borders Fig. 5, Plate 3 

DEFINITIONS. 

Lines and angles Text opposite Plate 2 

Planes and surfaces '' " Plate 3 

Polygons " " Plate 3 

Circle " " Plate S 

CONSTRCCnON OF PERPENDICULARS. 

To construct perpendiculars Nos. 1 to 6, Figs. 1 to 6. Plate 4 

To construct perpendiculars Nos. 7 and 8, Figs. 1 and 2, Plate 5 

DIVISIONS OF LINES. 

To divide lines Nos. 8 to 12, Figs. 3 to 6, Plate 5 

SOLUTIONS OF ANGLES. 

To construct and divide angles Nos. 13 to 16, Figs. 1 to 4, Plate 6 

SOLUTIONS OF TRIANGLES. 

To construct triangles Nos. 17 to 30, Figs. B and 6, Plate 6 and Figs. 1 and 2, Plate 7 

PROPORTIONAL LINES. 

To construct proportional lines Nos. 21 to 34, Figs. 3 to 6, Plate 7 

POLYGONS. 

To construct regular polygons on given bases Nos. 25 to S3, Figs. 1 to 4, Plate 8 

To construct equivalent polygons Nos. 33 to 40, Figs. 5 and 6, Plate 8 and Figs. 1 to 6, Plate 9 

TRANSFERRING POLYGONS. 

To construct polygons equal to a given one Nos. 41 to 46, Figs. 1 to 6, Plate 10 and Figs. 1 to 6, Plate 11 

REDUCTION OR ENLARGEMENT OP POLYGONS. 

To construct polygons proportional in area or outline and 

similar to given ones Nos. 47 to 53, Figs. 1 to 11, Plate 12 and Figs. 1 to 5, Plate 13 

SCALES. 

To construct scales Nos. 53 to 56, Figs. 6 and 6 A, Plato 13 and Figs. 1 to B, Plate 14 

CIRCLES. 

To divide circles Nos. 57 to 60, Figs. 1 to 5, Plate IB 

RECTIFICATION OP ARCS. 

To rectify arcs Nos. 61 and 63, Fig. 6, Plate 15 and Fig. 1, Plate 16 



INDEX.— Continued 

TANGENTS. 

To construct tangents Nos. 63 to 66, Figs. 2 to 5, Plate 16 

TANGENTIAL CIHCLES. 

To construct tangential circles Nos. 67 to 81, Figs. I to 6, Plate 17, Figs. 1 to 6, Plate 18, Figs. 1 to 4, Plate 19 

GOTHIC AND PERSIAN ARCHES. 

To construct Gothic and Persian arches Nos. 83 to 87, Figs. B and 6, Plate 19, Figs. 1 to 4, Plate 20 

EGG-LINES. 

To construct egg-lines Nos. 88 and 89, Figs. 5 and 6, Plate 20 

OVALS. 

To construct ovals Nos. nO to 93, Figs. 1 to 4, Plate 21 

ARCHES. 

To construct arches Nos. 94 to 98, Figs. 5 and 6, Plats 81, Pigs. 1, 2 and 4, Plate 22 

ASCENDING ARCHES. 

To construct ascending arches Nos. 99 to 103, Figs. 3 and 5, Plate 22, Figs. 2 and 3, Plate 83 

SPIRALS. 

To construct spirals Nos. 103 to 100, Fig. 1, Plate 23, Figs. 1 to 3, Plate 24 

CAM LINES OR ARCHIMEDEAN SPIRALS. 

To construct cams Nos. 107 to 110, Figs. 1 to 4, Plate 25 

CONIC SECTIONS— ELLIPSE. 

To construct ellipses Nos. Ill to 120, Figs. Sand 6, Plate 35, Figs. 1 to 6, Plate 26 

PARABOLA. 

To construct parabolas Nos. 131 to 130, Figs. 1 to 6, Plate 27 

HYPERBOLA. 

To construct hyperbolas Nos. 137 to 139, Figs. 1 to 3, Plate 28 

GEAR LIHES— CYCLOID AND EVOLUTE. 

To construct an evolute No 130, Fig. 4, Plate 28 

To construct a oyloid Nos. 131 to 133, Figs. 1 and 2, Plate 29 

To construct an epicycloid and hypocycloid (epitrochoid and 

hypotrochoid) Nos. 134 to 138, Figs. 1 to 3, Plate 30, Figs. 1 and 3, Plate 31 

APPLIED CONSTRUCTIONS IN ARCHITECTURE AND MECHANICS. 

To construct an ornamented Gothic arch No. 139, Fig. 1, Plate 32 

To construct a pair of "spur" wheels No. 140, Fig. 1, Plate 33 



Necessary Tools, Implements and Their Application. 



Fig. 1, Plate?. — A drawiiig-board, made of well- 
seasoned white pine, poplar (whitewood) or 
basswood, the lightest of our woods, answers 
this purpose best, as these woods are evenly 
grained and do not offer great obstruction to 
thumb-tacks, by which the drawing-paper is 
fastened to the board. 

The under surface of this board should be 
provided with two parallel dovetailed grooves, 
3 or 4 inches from edges O and C and right- 
angled to the grain of the wood, to receive not 
too tightly fitting cleats, at which the board may 
shrink, to prevent its splitting. The cleats 
therefore should not be glued in the grooves to 
receive them. 

When one draws with the right hand, the 
straightedge, called T square (T), and triangle 
S, called set square, are operated with the left 
hand, and when one draws with the left hand 
the set and T square are operated with the right 
hand. 

The T is used only on one edge of the hoard. 

Figs. 1 and 2, Plate 3.— Set squares (Triangles). — 
One set square of 30" and 60" and one of 45" 
(degrees) are required, as shown in Plate 3 and 
these should be tested for accuracy before ad- 
mitted to practical use. 



Test. — Place the set square with one right- 
angle side to the T, as shown in Fig. 1, and 
draw with a hard ( 4 H) well-pointed lead pencil 
a line on side a b. Reverse the set square on 
a b as an axis, and if the line drawn and the 
side of the set square coincide (fall into one) 
the angle is a right angle, while a convergence 
will show the angle to be incorrect, and such a 
set square should not be used until it is made 
tnie. A similar test should also be made with 
the T square before using it. 
Figs. 3 and 4.— Plate 3.— Figs. 3 and 4 show the 
different angles possible to be drawn with the 
assistance of both set squares and the T. 

Fig. 3, Plate 1. — The protractor is a semi-circular 
instrument made of brass or celluloid. Point 
C represents the center of the semicircle, which 
is divided by radii into 180 equal parts. In 
measuring an angle, such as the angle BOA, 
place the instrument with its center at the vertex 
(the intersection of the sides of the angle), and 
one side to coincide with the diameter of the 
instrument. Note the number of divisions on 
the intervening arc, which is 137 (read 137" 
degrees); 1" = 60m. (minutes) and 1 m. =60 sec. 
(seconds). 



Necessary Tools, Implements and Their Application. — Continued. 



THE set op drawing INSTRUMENTS. 

Figs. 4 to 7, Plate 1. — The very best is none too good. 
A set should contain one pair of compasses, 
Fig. 4, with needle-point center. Fig. 4 D, a lead 
pencil attachment. Pig. 4 B, a ruling-pen for 
circles, Fig. 4 A, one pair of dividers. Pig. 5, 
and one or two straight ruling-pens. Fig. 6, of 
different sizes. For boilermakers, machinists, 
architectural iron constructors, etc., a set of 
bow instruments is a valuable addition to the 
above. 

Pigs. 7 and 7a, Plate l.—T/ie lead (6 H) for the 
compasses is bought in sticks of 5 in. in length 
and j'g in. thick. Take a length i in. longer 
than the length of the hole in the attachment to 
receive it. Give the lead the shape shown in 
Fig. 7, which is most conveniently done on a 
piece of emery paper or a fine file ; then take 
off the corners as indicated by the lines G K 
and H I Fig. 7 A. Insert it with a iiat side 
towards the center of the compasses and clamp 
it tight with the clampscrew S, Fig. 4 B. 

The main joint near the handle ought to 
move with ease, and one hand should be suffi- 
cient to open or close dividers or compasses 
easily. 

The straight pen and the pen for circular 
ruling must be treated most carefully. Their 
blades are of the same length, not so pointed 
and sharp as to cut the paper, and when filled 
with ink should be entirely free of ink on the 
outside. 

In inking circles, the legs of the compasses 
should be bent at the joints P and O (Fig. 4) 



sufficiently to have both blades touch paper 
equally to allow an even flow of the ink. The 
leg which carries the center of the compasses 
should have a vertical position so as to avoid 
a tapering of the hole in the paper by its revo- 
lution. 

The correct position of the compasses is 
shown in Fig. 4, where the line M N represents 
the surface of the drawing-paper. A convenient 
arrangement to keep the plates of the course 
for later reference is used in the Chicago High 
Schools. The sheets of paper of 11 in. X 17 in. 
are perforated and seamed by laces in a porte- 
folio of 13 in. X 18 in. A rectangle as a bor- 
der line of 10 in. X 15 in. encloses the drawing 
surface which is divided into six equal squares 
5 in. on a side, each to receive one construc- 
tion. See Plate 4. Larger spaces however 
should be used to execute accurately some of 
the constructions, for which the proportional 
sizes may be ascertained from the correspond- 
ing plates. Draw the lines light and carefully 
withDixon'a VH (very hard), Faber4H (Sibe- 
rian), or a Hartmuth 6 H (compressed lead) 
pencil, having a fine round point. 

Inking the drawing. — Execute all construc- 
tions in pencil, to admit of corrections, when 
necessary, before inking them. It is also ad- 
visable for the inexperienced to write the re- 
quired text on the drawing in pencil, to distri- 
bute letters and words regularly in the avail- 
able space beneath each drawing, as shown in 
Fig. 1, Plate 4, before writing with Indian ink. 



Plate ± 




Hnmtein's Constniclivc Drawinij, 



Plates S and B.—Alpliahets. — Several styles of lettering for 
titles of drawings commonly used, are shown on Plates 
2 and 3. 

Plate 3 —In Fig. 5, ABCBEFG and H show a few samples 
of corners in border lines for elaborate work. 

The following distinctions of inked lines in drawing are made 
to recognize readily all that pertains to proJnem, construction 
and res u it. 

Thb problem line is drawn fine and uninterrupted. 

The construction line is fine and dashed. 

The result, a strong, U7iinterrupted line. 

Begin inking with construction arcs and circles, then the cir- 
cular problem lines, and then the circular result lines. 

This is done so as to save time, to avoid the change of tool in 
hand , and not to clean and rs-set the pen of teuer than necessary. 

Constritction t-traight lines are drawn next very fine and 
dashed, corresponding to construction arcs and circles, and 
lastly the 

Result straight line, to corrgspond to result arcs and circles. 

The inking of a drawins is a recapitulation of each construc- 
tion, and this important work should be executed with great 
care. 



A postulate is a statement that sora,ething can be done, and is 
so evidently true as to require no reasoning to show that it can 
be done. 

An axiom is a truth gained by experience, and requiring no 
logical demonstration. 

A theorem is a truth requiring demonstration. 



LINES AND angles. 

A right line is the shortest distance between two points. 

When the term line alone is used, it indicates a right line, A 
veitical line is the *' plumb-line" ; a horizontalUne, one making 
a right angle with the vertical and a line of any other direction» 
is called oblique. 

A curbed line or curve changes its direction in every point. 

When two lines lying in one plane, on being produced in either 
direction, do not intersect, these lines are said to be parallel. 

Two lines in one plane, which have a diSerence of direction, 
are said to form an angle. The point of intersection of these 
two lines, called sides, is the vertex of the angle. 

When two such lines intersect each other, so that all four 
angles formed are equal, we say they are right angles. The 
common vertez:of these four right angles may be assumed to 
be the center of a circle, v/hich by diameters is divided into 3G0 
equal parts, called degrees (°). Each angle contains % of 360"^ 
= 90°, which is the right angle. An angle gi-eater than 90° is an 
obtuse angle; &!! angle smaller than 60° \qw[i acute angle. When 
the two sides of an angle form a straight hne, the angle is called 
a straight angle, and its magnitude is 180°. 

Generally we desigTiate an angle by three letters, for instanco, 
h ae or cab; then the middle letter (a) indicates the vertex, 
while the sides are b a and c a. 



Plate 2 



E^igineering Script, (Rotnan) 
120*5 abcdeFghijhl'm''}topcrrstuvivxyz. ejeao. 

ABCDEFGHIJhLMJ\'OPQRST[fVWXYZ. 

I2^i-^ abcdefg^ijKlnAiAopcj-KstuVwXijz. 67590. 

Shop Skeleton. 
abcdefAhij kimn opcfKstu vwxyz. 

I234S ABCDEFGHIJKLMNOPQRSTUVWXYZ. 67330. 

AB CDEFGHIJ KLMNOPQ RSTUVWKYZ 

dVcmW) - ZA'X'i/tt/wa'. 

/af^c-i^-^zJA^ lliiftl m nop cj^<it^^^'^u %U2. 123^567890. ^ '^ 



Hmutcin's Con«tnic(i04 Draun'nff. 



AND SURFACES. 

A plane has two dimensions— length and breadth, 

A surface is the boundary of a body. 

Surfaces bounded by right Unes are called polygoiis. Regular 
polygons have equal sides and equal aiigles; they are equilateral 
and equiangular* 

POLYGONS are: 

The triangle, which has 3 sides, 

"■ tetragon or quadrilateral, " 

" pentagon, ** 



heptagon, 

octagon, 

enneagon or nonagon, 

decagon, 

undecagon, 

dodecagon. 



12 



etc. 



The triangles are: The equilateral triangle which is also 
equiangular; the isosceles triangle, having two sides equal, and 
the scalene triangle, whose sides are unequal. 

An obtuse and a right-angled triangle have one obtuse and 
one right angle respectively. An acute angled triangle has 
three acute angles. The side or *'leg" opposite the right angle 
in a right-angled triangle is called the hypotenuse, the sides or 
legs forming the right angle are the catheti. 

The sum of the squares constructed 07i the catheti is equivalent 
to the square erected on the hypotenuse. ' 

The sum of all angles in a triangle is equal to two right 
angles. 

A line drawn from a Tertex of a triangle perpendicular to 
the opposite or produced opposite side is called its altitude or 
heiaht. 



quadrilaterals. 

The square has equal sides and 4 right angles. 

The rectangle has opposite sides equal and 4 right angles. 

The rhombus has equal sides and equal opposite angles. 

The trapezoid has only two parallel sides. 

The trapezium is an entirely irregular quadrilateral. 

Quadrilaterals which have the opposite sides parallel are 
parallelograms. 



Fig. I. Plate 15.— Definition.— A circle is a curve the points 
of which are equally distant from a fixed point, called the 

center. 

The distance from the center to any point of the circle is 
called the radius. The connecting line of any two points of the 
circle is called a chord. If the chord is produced to any point 
outside the circle, it is called a secant. The chord through the 
center is called the diameter. The circle considered as a length, 
is called a circumference. Any arbitrary part of the circum- 
ference is called an arc. The arc that forms the fourth part of 
the circumference is called a quadrant; the sixth part a sex- 
tant; the eighth part an octant; while half the circumference 
is called a semi-circle. The area comprised by two radii and 
the intervening arc is called sector; the area comprised by a 
chord and the corresponding arc is called a segment of a circle. 

In Fig. 1, C D, C B and A are radii, G H is a chord, E I F is 
a secant, A B is a diameter, Q J H an arc, area D O B L D is a 
sector, area H G J H a segment, tract A J D B a semi-circle. 

Postulate. — Draw a circle, if the center and the radius are 
given 



Plate 3. 



BLOCK. 



Sr i:iMI 



EFGHIJKLMNOPOR 

STUYWXYZ. 1234567890. &. 

^/i^iJjj;iLj\Jj^JDj^-g;] 



rtrr- 




3-iclL S"t,q.2. S\q\ 



Haimtein's-ConsUiictivc Dratptno, 



CONSTRUCTIONS. 



1. — Fig. 1. — Problem. — At a given point in a given 
line to erect a perpendicular, or to bisect a straight 
angle. 

Solution. — Let M N be the given line and A the 
given footpoint of a perpendiculai". From A as a 
center and with any radius describe the circle B, C; 
B and C ai'e equidistant from A and are the centers 
of arcs with equal radius greater than B A, which 
intersect at point D. Draw the line D A, which is 
perpendicular to the line M N, in point A. 

2. — Fig. 2. — Probiem. — To draw from a given point 
a perpendicular to a given line. 

Solution. — With the given point A as a center 
describe a circle intersecting the given line M N in 
two points, B and C. From B and C as centers and 
with equal radii draw arcs intersecting at D. Con- 
nect points A and D by the line A D, which is per- 
pendicular to M N. 

3.— Figs. 3, 4, 5 and 6. — Problem. — To erect a per- 
pendicular at the end of a given line, M A. 

Solution. — Take any point C outside of M A as a 
center, and with a radius C A describe a circle in- 



tersecting MA at D. Draw the diameter D C B. 
Connect points B and A by the line B A, which is 
the perpendicular to M A. 

4. — Fig. 4. — Solution. — From A as a center and any 
radius describe the circle B N, at which make 
B C = A B and pass through points B and C 
the line B C indefinite; make then C D =' C B 
and connect A and D by the line A D, which is 
the required perpendicular. 

5. — Fig. 5. — Solution. — Describe from A as a center 
and any radius the circle B C E. Make E C = 
C B = B A, and from E and C as centers and 
with equal radii draw intersecting arcs at D. 
Connect D with A with a line, and D A is the 
required perpendicular. 

6. — Fig. 6. — Solution. — From A towards M lay down 
a division of 5 equal units. With A as a cen- 
ter and 3 units as a radius draw the arc 3 B in- 
definite, and 4 as center and 5 units as radius 
cut the arc at B. Connect B with A, and line 
B A is the required perpendicular. 















Plate 4. 


Fig. 1. 




Fig. 2. 


Fig. 3. 


/ 










/ N 

/ \ 


\ 


-" 


1 




„ 3 


M dX^ 


^ 


/Cll'<iaiVen«omtm aaLvervflnetO 








F':g'. ■*■ 




Fig. 5. 


Fig. 6. 


/ / 

.... . . c 


) 


/ 
/ 

1 
1 

Bl 


~^-s 




B 


M B, 






H i r L i 1 





Hansteln'^ ConstnicUve Drawing. 



7. — Fig. 1.— ProWsm, — To construct a perpendicular 
at or near to the end of a given line. 

Solution. — When M N is the given line, take in 
MN an arbitrary point A as a center and a radius 
longer than A N ; describe arc C E D. From an 
other point, B, near N, with any radius, draw arcs 
intersecting circle (A) at C and D. Connecting 
points C and D by a line we have the required per- 
pendicular. 

DIVISION OF LESTES. 

8. — Fig. 2.— Problem — To hisect a line. 

Solution. — When A B is the given line, use A as 
center, and with a radius greater than i A B draw 
the arc D C E. With the same radius and center 
B draw an arc to intersect the arc D C E in points 
D and E, which are connected by the line D E. The 
line D E will not alone cut the line A B into two 
equal parts, but will also be a perpendicular to 
AB. 

Figs. 8, 4, 5 and 6.— Problem. — To cut a given line 
into any number of equal or proportional parts. 

9. — Fig. 3. — Problem.— 4 line A B shall be divided 
into 7 equal parts. 

Solution.— Draw the line B N at about 35° to A B. 
Lay thereon, starting from B, seven times a unit 
and connect points 7 and A by line 7 A. Parallel 
with line 7 A draw lines from each division point, 
6, 5, 4, etc., which will divide A B into the required 
number of 7 equal parts. A C is y of A B. 



Remark. — ^Parallel lines are drawn with the set 
and T square combined. Adjust the longest side of 
the set square to coincide with the line with which 
we intend to draw .parallels, and place the T to one 
of the right- angle sides of the set square. Keep T 
firmly in this position and slide the set square 
along its edge in the required direction and draw 
the parallels. 

10.— Pig. 4.— Problem.— To cut a given line into two 
proportional parts, as 3:8. 
Solution. — Draw the line B N, and from B lay 
down a division of 3+8 equal parts. Connect 
points A and 11 by the line A 11 and draw parallel 
with it 3 C. C B is ^^ and C A ^^ of A B. 

11. — Fig. 5. — Problem. — To cut a given line into three 
proportional parts, as 7:3J:lf. 
Solution. — Draw the line B N, and from B lay 
down a division of 7+31+11 equal parts. Connect 
point 12J with A and draw parallel with 12J A, the 
lines lOi C and 7 D. A C is then li, C D 3i, and 
D B 7 parts of line A B. 

12. — Fig. 6. — Problem. — To cut a given line into any 
number of equal parts by a scale. 
Solution. — Draw a rectangle A 14 N, and divide 
A N by horizontals into any number of equal 
parts, and number them 0, 1, 3, 3, 4, 5, etc. When, 
f . i., the line A B is to be divided into 9 equal parts, 
take the line to be divided as a radius and A as 
center; describe an arc to intersect line 9 at point 
B 9, which connect with A by line B 9 A. By the 
horizontals the line B 9 A is divided into 9 equal 
parts. In the same figure the problem is solved to 
divide the line A B into Hi equal parts. 



c 

/ 


v" 




/ 



Plate 5. 

Pig-3.\ 



\ \ \ \ \ V 
\ \ \ \ \ V^ 

\ \ \ v^ 
^ A, 



F;g-. 4. 



Fig. 5. 



Fig. S. 



I 



^ ^^ 







Hanstcin's Constructive Draifing. 



SOLUTION OP ANGLES. 

13. — Fig. 1.— Problem. — To construct an angle equal 
to a given one. 

Solution. — Angle C A B is the given angle. When 
the vertex O and one side O N of the angle to be 
constructed are given, describe with O as a center 
and A C as a radius the arc E D, and from D as a 
center with the radius B C the arc at E ; draw the 
line E O. Angle C A B = angle E O D. 

14.— Fig. 3.— Problem — To bisect an angle. 

Solution. — Let B A C be the given angle. With 
A as a center and a radius A B draw the are B C. 
B and C are the centers for arcs with equal radii, 
intersecting at D ; draw line D A, which divides 
BAG into two equal parts. 

15.— Pig. 3.— Problem — To trisect a right angle. 

Solution. — From vertex A, with the radius A B, 
draw the arc B C. With B as center and the same 
radius draw the arc A E, and from C the arc A D; 
draw lines D A and E A. Angle BAD = DAE = 
EAC. 

16.— Fig. 4. —Problem.— To trisect any angle. 

Solution by Dr. Henry Eggers. — Let C A B be 
the angle to be trisected. Describe with A as center 
a semicircle BCD, which intersects the prolonged 



side B A of tbe angle at D ; draw from C an arbit- 
rary line OEM and make E F = E A, and draw 
F G C; then make G H = G A and draw H I C. An 
additional operation will not be necessary, as the 
lines will fall so close together as to almost coin- 
cide, and it is angle C H B which equals J C A B. 
This construction is convenient for angles up to 
90°; and in case of the trisection of an obtuse angle 
we bisect first and then trisect, so that the double 
third of the bisected angle is equal to the third of 
the given obtuse angle. 

SOLUTION OP triangles. 

17. — Fig. 5. — Problem. — To constmct a triangle when 
the three sides are given. 

Solution. — Lines 1, 2 and 3 are the sides given. 
Lay down line B C = line 1. Prom C as center, 
with line 3 as a radius, draw an arc, and with line 
3 as radius and center B another arc, intersecting 
the first arc at D. Draw lines D C and D B ; then 
D C B is the required triangle. 

18.— Fig. 6.— Problem.— To construct a triangle of 
ivhich two sides and the included angle are given. 

Solution. — Construct angle D, and from its vertex 
cut off the sides 1 and 3, that is C B and C E, and 
draw line E B; then E B C is the required triangle. 




Fig. 2. 




Pla,te 6. 

Fig. 3. 



—- 


-^/ 


', 


/ 

/ / 
/ / 
/ / 
/ / 






C^-^-^ 









Fig. G. 




Hansfein's Constructive Drawing. 



19.— Fig. 1.— Problem.— To construct a triangle of 
which one side (1) and the two adjacent angles D 
and E are given. 

Solution. — Lay off C B equal to linel; transfer 
the angles D and E on line C B, and prolong the 
sides to intersect at F; then triangle C F B is the 
required triangle. 

20.— Fig. 2. — Problem. — To construct a triangle of 
which one side (1), one adjacent angle D and one 
opposiite angle E are given. 

Solution. — Construct C B equal line(l) and angle 
D at C as before ; at an arbitrary point E on line 
C M draw angle C M N = E, and parallel with M N 
the line B F. F is the third vertex of the required 
triangle C F B. 

PROPORTIONALi LINES. 

21.— Fig. 3.— Problem.— To construct to three given 
lines a fourth proportional. 

Solution. — Lay down an angle M A N of about 
40", and from A cut the segments A 1 ^ line 1, A 2 
= line 2, A 3 = line 3 ; draw line 2 1, and with it 
parallel the line 3 x. A x is the required line. 
1 : 2 = 3 : A X. 



22.— Pig. 4.— ProbJem —To construct to two given 
luies a third proportional. 

Solution. — Lay down the angle as before, and 
from A cut the segments A 1 ^ line 1, A 2 = line 2, 
A 2' = line 2. Draw line 2 1, and parallel with it 
2' X. Ax is the required line. 1:2 = 2:A x. 

23. — Pig. 5. — Problem. — To construct a mean pro- 
portional to two given lines. 

Solution. — A B + B C is the sum of the given 
lines 1 + 2. Find point D, the center of A C, and 
a radius D A; draw the semi-circle A X C. Erect 
at B a perpendicular B X, which is the required 
line. AB:BX=BX:BC. 

24. — Fig. 6.— Problem To construct to a given line 

major and minor extreme proportionals. 

Solution. — At point B of the given line A B erect 
a perpendicular B D = i A B, and draw line ADP 
indefinite; with D as center, D B as radius, describe 
semicircle EBP, and from A as center, A E as a 
radius, draw arc E X. The line A F : A B ^ 
A B : A X. 



Plate 7. 




Fig. 2. 




\ \ 



Fig. 5. 



\ \ 
\ \ 
\ \ 

i ^\- 

2 I 





^ 


■^ 1 


r^^ 




/ 






/ 

/ 

/ 


1 


\ 


/ 




1 






Harniitin"^ ConstruclAve Diaauing. 



POLYGONS. 

25, — ^FiG. !•— Problem. — To construct a regular triangle on ^ 
given ba^e. 

Solution.— A B is the given base. "With A and B as centers and 
A B as radius draw arcs intersecting at G. Draw the lines C A 
and C B. A C B Is the required regular triangle. 
26.— Fig. 1.— Problem,— To construct a regular hexagon on a 
given base. 

Solution. — Let A B be the given base. Construct on this a 
regular (or equilateral) triangle. The vertex C is the center, 
and C A = C B the radius of a circle, in which a regular hexa- 
gon A B D E F G, with A B as side, can be inscribed. 

Corollary. — A regular hexagon may be divided into six equal 
equilateral triangles, the common vertices of which lie in the 
center of it. 

37.— Fig. 3.— Problem.— To construct a regular heptagon at a 
given base. 

Solution. — Draw with the given base A B the equilateral tri- 
angle A 6 B, as in the previous construction. From center D of 
A B draw the line D 6 13 perpendicular to A B. Divide 6 A into 
six equal parts. These parts transfer on line 6-13 and number 
them, 7, 8, 9, 10, 11 and 13. Point 7 is the center, and 7 A the 
radius of a circle, in which the regular heptagon A B C D E F G, 
with A B as side, can be inscribed. 

28.— Fig. 2.— Problem.— To construct a regular polygon with 
more than 6 sides. 

Solution.— With points 7, 8, 9, 10, 11 and 13 as centers, and 7 A, 
8 A, 9 A, 10 A, U A and 13 A, respectively as radii, draw circles 
in which the line A B as repeated chord will form the regular 
heptagon, octagon, enneagon, decagon, undecagon and dode- 
cagon. 

RemarU. — Regular polygons with greater number of sides are 
rarely used in practice, and are therefore omitted here. 
29.— Fig. 3.— Problem.— To construct a square at a given base. 

Solution.— liet A B be the given base. Draw at A and Q per- 
pendiculars with set and T square, and make A G = A B, and 
with T square draw CD. A O D B is the required square. 
30,— Fig. 3.—Problem.— To construct a regular octagon at a 
given base. 

Solution.— In the bisecting point H of the given base A B 
erect a perpendicular. H F, at which make H E -= A H and E F 
= E A. F is the center and F A the radius of a circle, in which 
draw A B eight times, as repeated chord, to complete the re- 
quired octagon ABGHIJKL. 



31.— Fig. 4.— Problem,— To construct a regular pentagon at a 
given base. 
Solution. — Let A B be the given base; produce it towards N. 
Erect at B a perpendicular, B D = A B. Bisect A B by point C; 
with G as center and G D as radius draw arc D E. With A and 
B as centers and A E as radius draw arcs to intersect atF. 
With F and A as centers draw arcs intersecting at G; and from 
F and B as centers, with the same radius A B, draw arcs inter- 
secting at H. Connecting B H, H F, F G and G A by lines we 
complete the required pentagon A B H F G, 

33.— Fig. 4.— Problem,— To construct a regular decagon at a 
given base. 

Solution.— L.6t A B be the given base. Follow the construc- 
tion of the pentagon until the position of point F is found; this 
is the center, and F A the radius of the circle, in which as re- 
peated chord the line A B will complete the required regular 
decagon ABIJKLMNOP. 

33.— Fig. 5.— Problem,— To consttmct tHangles equivalent to a 
given one . 

Solution.— liet A C B be the given triangle; draw line M N 
parallel with A B through point G. Locate an arbitrary point 
E or G in line M N, and draw lines E A and E B, and G A and 
G B. Triangle A E B = A C B = A G B. If one side of the tri- 
angle is called the base, a perpendicular drawn from the oppo- 
site vertex to the base, or produced base, is the altitude or 
height of the triangle, as B F, C D and G H. 

Theorem. — Triangles of equal base and altitude are equi- 
valent . 

34.— Fig. 5, A.— Problem.— To construct parallelograms equiva- 
lent to a given one. 

Solution. — Let A B D C be the given parallelogram, with base 
A B. Draw the line M N parallel with A B, make E F and G H 
= C D, and draw lines E A. F B, G A and H B. The parallelo- 
gram EFBA==CDBA = QHBA. 

In a polygon any right line which passes through two non- 
consecutive vertices of its circumferential angles is called a 
diagonal, 

ITieorem.- Either diagonal divides the parallelogram into two 
equal triangles. 

85.— Fig. 6. — Problem.— To construct a rectangle equivalent to a 
giveyi triangle. 

Solution,~A B C may be the given triangle, and C F its alti- 
tude. Bisect C F rightangularly by line D E, and erect the per- 
pendiculars B E and AD. A D E B is the required rectangle. 



Fig. 1. 





Plate 8. 

Fig. 3. 



-^ 


1 \ \ 


l.^rf,'^ \ 


r/ 1 \ 


i/ \« 



A — — — B 



Fig (7. 





Fig. 5. a. 







Ham&uin's Constniclive Drawing. 



36. — Fig. 1. — Pfoblem — To construct a rectangle 
equivalent to a given trapezoid. 
Solution. — Let A B C D be the given trapezoid. 
Bisect rightangularly its altitude L M by tiie line 
I K, which bisects also the sides B A and C D in I 
and K. Perpendicular to I K, through I and K, 
draw P G and EH. F E H G is the required rect- 
angle, equivalent to the trapezoid A B C D. 
37.— Fig. 2.~ProbIsEa. — Tlie side of a square is given: 
to construct the sides of squares that are twice, 
three times, four times, etc., as great as the square 
over the given line. 
Solution. — Construct a right angle B A 1 ; make 
B A and A 1 equal to the given side of the square; 
then lay off successively A2 = B1,A3 = B2, A4 
= B 3, etc. A 2, A3, A 4, etc., are the sides of 
squares that are respectively twice, three times, 
four times, etc., the area of the square over A 1. 
38.— Pig. 3.— Problem To construct a tria7igle equi- 
valent to a given irregular pentagon. 
Solution. — Let A B C D E be the irregular penta- 
gon. By the diagnols A C and C E divide it into 
three triangles A B C, C A E and ODE. Produce 
the base A E to the left and right indefinitely, and 
parallel to C A draw the line B F ; connect C with 
P ; then draw D G parallel with C E and connect C 
with G. The sum of the triangles C F A + C A E 
+ C E G is equal to the triangle CFG, which is 
equivalent to the irregular pentagon ABODE. 



39. — Fig. 4. — Problem — To construct a square equi- 
valent to a given triangle. 

Solution. — Let CFG, Pig. 3, be the given tri- 
angle. Construct a mean proportional between 
half the base F G and altitude C H, as shown in 
Pig. 5, Plate 7, by making I K =- 4 P G, and K L = 
C H. The sum I K + K L is the diameter of the 
semicircle I N L. Erect at K a perpendicular, which 
is intersected by the circle in N. N K is the re- 
quired side of the square, and NOPK is the square, 
which is equivalent to the triangle CFG and the 
irregular pentagon A B C D E. 

40. — Figs. 5 and 6.— ProbSern. — To transform an ir- 
regular heptagon into an equivalent triangle and 
square. 

Solution. — Let ABCDEFG be the irregular 
heptagon. Draw line C A, and parallel to it B N; 
connect N and C by line N C. Triangle C N A = 
C B A. Treat the triangle E F G in a similar way, 
and you have transformed the heptagon into the 
irregular pentagon N C D E M. Proceed as in Fig. 
10, and transform the pentagon into the triangle 
D H I; transform this into the square P Q R O, 
which then is equivalent to the given heptagon 
ABCDEFG. 



C E 

a. 



Fiff. 




Plate 9. 
Fig. 3. 




Fig. 4. 





H N L M G 1 




HarvsUin's Constntctive Diav:in'j. 



TO TRANSFER POLYGONS. 

41. — Figs. 1 and 2.— Problem.— To construct a poly- 
gon equal to a given one by triangles. 

Solution. — Polygon Fig. 1 is given and divided 
into triangles, ABC, etc., etc. Draw line A' B' 
parallel and equal to A B. On A' B' construct tlie 
triangle A' B' C equal to triangle ABC. The 
remaining triangles of Fig. 1 lay off in the same 
order and position as Fig. 2 shows, starting from 
side B Cj; then polygon (Fig. 2) is the required 
one. 

42.— Figs. 3 and 4.— Problem,— To construct a poly- 
gon equal to a given one hy sectors. 
Solution. — Polygon Fig. 3 is given. From center 
O with any radius describe circle C B D, etc., and 
draw from center O a radius to each vertex of the 
polygon to intersect with the circle. Locate center 
C, Fig. 4, and with radius C D' equal O D describe 
the circle D' B' C, etc., and draw C D' parallel 
to O D; make arcs D' B' = D B, B' C = B C, etc., and 
pass lines through points D', A', C, etc.; further 
make O' E' = OE, O' A' = A, C F' = OF, etc., 
and by connecting points B' A' F', etc., complete 
the required polygon. 



43.— Figs. 5 and 6.— Problem.— To construct a poly 
gon equal to a given one, by co-ordinates. 

Semarlc. — In the plane of drawing a convenient 
line is drawn (horizontal), called the axis of ab- 
scissae; the position of the different vertices of the 
given figure is determined by perpendiculars, called 
ordinates, from these vertices to the axis of 
abscissae. Take any convenient point, A, on this 
axis and draw a perpendicular to it, M N. This 
line is called the axis of ordinates, and reckoned 
fi'om this point A (called the origin) the segments 
determined by the loot-points of the ordinates are 
called abscissae. The common appellation of both 
systems of lines (the abscissae and ordinates) is 
co-ordinates. 

Solution. — Pig. 5 is the given polygon. Through 
any vertex (origin) draw a horizontal, A R, and 
perpendicular to it the ordinates from each vertex 
or principal point for transmission. Draw A R', 
Pig. 6, and lay off A B, AC, AD, etc., = AB, AC, 
A D, etc., of Pig. 5. Erect the perpendiculars A A', 
B B', C C C" C", etc., and make A A', B B', C C, 
C C", C C", etc., equal to the corresponding per- 
pendiculars in Pig. 5. Connect A and A', A' and 
B', describe with radius C C, center C, arc B' C", 
etc., and complete the requii-ed polygon. Pig. 6. 




Fig. 2. 




Plate 10. 

F'S- 5. 




Fig. 3. 



Fig. 4. 



Fig. 6. 






Haiistein's Consttilctive Drawing. 



44.— Figs. 1 and 2.— Problem — To construct a poly- 
gon equal to a given one, radiating in a circle. 

Solution. — ^Let A E D G, etc., be the given poly- 
gon. Describe with A D, A E, etc., as radii and A 
as center the circles C D, E F G, etc., and make 
F' E' = F E, F' G' = F G, etc. Connect D' and E', 
D' and G', etc., and complete the required polygon. 
Fig. 2 shows the construction applied to other 
polygons. 

Bemark. — This construction is used conveniently 
to draw a rosette in which an ornamental unit 
occupies a sector division of a circle. 

45.— Figs. 3 and 4.— Problem — 2b construct sym- 
metric polygons or outlines. 
Solution. — Let L M, etc., be the given outline as 
a profile of the base of a column. Draw the hori- 
zontals L L', M M', etc., and the axis of symmetry 
R N. Make A L' = A L, B M' = B M, D O' == D O, 



etc. Connect L' and M', etc., and complete the re- 
quired symmetric profile of the base of the coluunn. 
46. — Pigs. 5 and 6. — Problem. — To construct an ir- 
regular outline equal to a given one. 
Solution. — Let B A C be the given outline. Cover 
this with a series of equal small squares and con- 
struct in Fig. 6 the same number of equal squares 
arranged as in Fig. 11, and transfer the points of 
intersections of the irregular outline with the sides 
of the squares ; make M' A' = M A of Fig 5, and 
M' B' = M B, etc. Connect B' A' C by a free-hand 
line and complete the required irregular outline, 
Fig. 6. 



Fig. 1. 




Plate 11. 

Fig- 5. 




Fig. 3. 




Fig. 4. 



Fig. G. 




Hartstein's Comtrttctivc Drawing. 



REDUCE OR ENLAH 



POLYGONS 



OUTLINE OR AREA. 



47.— Figs. 1, 2 and 3.— Problem.— To construct a polygon sim 
ilar to a given one of i its circumference. 

Solution.— liBtD ABC, etc., Fig. 1, be the given polygon. 
Construct the Scale Fig. 2. A perpendicular O 7, longer than 
the longest side of the fjiren polj-gon, is divided into 7 equal 
parts ; draw a horizontal line O N of an arbitrary length and 
■ connect points 7 and i with N by the lines 7 N and 4 N. O 4 is 
-1,-4 7 is s of the line O 7. All lines between O N and 7 N 
and parallel to O 7 are divided by 4 N and 7 N in the same pro- 
portion. To obtain the length of A' B', Fig. 3, place line A B in 
the scale as indicated by line A B' B, of which A B' is * of line 
A B. Transfer the remaining sides of the polygon by parallels 
and And of each the proportionate length in the scale Fig. 2, as 
shown by line A B; D' A' B' C, etc., is the required polygon. 

48.— Fias. 1, 4 and 5.— Problem.— To constriKt a polygoyi sim- 
ilwr to a given one, having 4 its area. 
Solution.— I^et D A B C, etc., Fig. 1, be the given polygon. On 
a horizontal line O 4 lay down a division of 7-|-4 equal parts and 
make O 4 the diameter of a semi-circle O M 4. Erect at point 7 
the perpendicular 7 M and draw lines M O and M 4. Then make 
line M B' equal to A B of the given polygon and draw B' B" 
parallel to O 4; A" B" (Fig. 5) — M B" in the scale Fig. 4. In 
relation to the side A B of the given polygon. A" B" is the side 
of a polygon, whose area is 4 of the given one. Treat the re- 
maining sides of the polygon similar to the side A B and com- 
plete the required polygon D" A" B" C", etc. 



49.— Fig. 6 —Problem.— To construct similar polygons which 
have I the circumference and a the area of a given one. 

Solution for circumference reduction.— Let D C B A E, etc., be 
the given polygon. From any point O therein draw radii to the 
vertices D C B A E, etc., and divide any one radius (O D) into 5 
equal parts. Parallel to D C from point 3 draw D'C, with 
B, O' B', etc., and D' C B' A E, etc., is the required polygon. 

Solution for area reduction.— Make radius O D the diameter 
of the semi-circle O N D and erect at division point 3 the per- 
pendicular 3 N and draw N O. Make O D" = O' N and proceed 
as before in drawing D" C" parallel with D C, C" B" with C B, 
etc. D" C" B" A" E", etc., is the required polygon. 

50. — Figs. 7, 8 and 9.— Problem.- To reduce any irregular out- 
line in proportion 8:5. 

Solution.— Let G H I K be the given irregular outline. Cover 
the given outline by a net of equal squares, the sides of which 
wo reduce by the scale. Fig. 8, to A' B' = ^a of A B. Draw with 
A' B' as unit the same number of squares as in Fig. 7. Transfer 
the points of intersection of the irregular outline with the sides 
of the squares, in reducing their distances from the vertices of 
the squares by scale Fig. 8, and transfer into Fig. 9. Connect 
these points by a free-hand Une, which is the required reduction 
of the irregular outline. 

Treat the surface reduction, Fig. 11, with the assistance of 
the scale Fig. 10 in a similar way, and we obtain the reduction 
in area. 



Plate 12. 



Fig.l 



Fig. 3. 



Fig. 5. 





( 




4L, 


^ 


\' 




) 








"[ 












) 








( 








)\ 


I 






















^ 









Fig. JO. 




Fig.li. 



^^t — rf 



Hanslctn's Cmistnictirj-- Dialling, 



51.— Figs, l, 2 and 3.— Problem — To construct a 
polygon similar to a given one, and of | its cir- 
cumference. { Transfer hy triangles. ) 

Solution. — Let A B D C, etc., be the given poly- 
gon. Construct the linear scale in proportion 2:3 
Fig. 2 similar to Fig. 2, Plate 12, and divide the 
given polygon by diagonals into triangles. Line 
A B' in the scale (Fig. 2) = A' B' of the polygon 
Fig. 3, whose circumference contains 3 units to 2 of 
the given polygon. Transfer and complete by tri^ 
angles the required polygon A' B' D' C, etc., Fig. 3. 

52.— Figs. 1, 4 and 5.— ProbJem,— To construct a 
polygon similar to a given one, which contains 3 to 
each 2 square units of the given polygon. 

Solution. — In the scale Fig. 4 the diameter of the 
semicircle consists of 3 + 3 equal parts; erect 3 M. 
Draw MO and M 3. Make M B' Fig. 4 = A B of 
the given polygon and draw B' B" parallel to O 3, 
A" B" = M B" of the polygon. Fig. 5, whose area 
has 3 square units to 3 of the given polygon. 

Transfer and complete by triangles the required 
polygon A" B" D" C", etc., Fig. 5. 



53.— Fig. 6.— Probiens.— To construct a scale of de- 
cimal division. 

Bemarh. — Small subdivisions of a unit which we 
cannot accurately perform with the dividers are 
constructed in Figs. 1 and 2. 

Solution. — Let line A, 7 = 8 centimeters, AO=AB 
= 1 cm. The decimal subdivision (millimeter, 
mm) is obtained by dividing O N into 5 equal parts 
by the horizontals in points A, B, C and D. Bisect 
B N and draw lines 5 A and 5 O ; line A 1 = ^oj 
B 2 = i^o, C 3 = 1%, etc., of O A, or 1, 2, 3, etc., mm. 
the required division. 

54.— Fig. 6 A.— Problem.- ro divide a centimeter 
into 100 equal parts. 

Solution. — Upon a straight line A 7 lay off eight 
units (cm) and construct squares on these distances. 
Let first square be A B N O. Divide sides A B and 
B N into ten equal parts (mm). Draw horizontals 
through division points on A B. R, being the 
first point of division from N to B, is connected 
with O, and through the other points parallels to 
R O are drawn between B N and A O. These 
parallels subdivide the millimeter (mm) into tenths- 



Plate 13. 





Fig. 6. 



S]Zx,t^x. . 



ig. o. A f^^iiiij,ctiou §-C/aC« -ill Sl^/jt^x .-i-,. 




' 8 DEC.M. 



Banxlein's Cnnstnictivc Drawitijg. 



55. — Fig. 1. ^Problem. — To construct a scale in which 
an inch is divided into 64ths. 

Solution. — Let A 2= 3 inches. Divide BN and 
B A into 8 equal parts each and complete the scale 
in the manner explained in Problem 54, Plate 13, 
Fig. 6 A. Line R O divides line R N = i in. into 
8 equal parts, hence into 64ths. Example : Take 
from this scale a line of 1|} inch (|| = J + a^). 
From O to 4 = f in.; follow the oblique line 
upward to the 5th horizontal point, n. Line N A = 
I in., A B = 6^ in- and b M = 1 inch and line N M = 
1|J in., as required. 

EEDUCTION SCALES. 

56.— Figs. 2 and 3.— Problem.— To construct a de- 
cimal reduction scale and draw by co-ordinates 
a polygon whose equations are indicated at tables 
A and B, Pig. 2 A. 



Remarh. — To draw the scale and polygon in con- 
venient proportion let the unit O A = 2J in., which 
may represent 100 feet. 

Solution. — Let O A be the unit to represent 100 ft. 
in the decimal reduction scale and let A 200 = 3 
such units. Divide O A, A B and B N into 10 equal 
parts, draw horizontals from 9, 8, 7, etc., and the 
oblique parallels with R O from division points 
10, 20, 30, etc., and we have the required decimal 
scale. Example: Take from this scale a line to 
represent 178 feet. Begin at point O, pass to the 
left to 70, then upward the oblique line to the third 
horizontal point R. Line R A =: 70 ft. A B = 8 ft. 
and B s = 100 ft, and ra + ab + bs = 173 feet. 

The polygon. Fig. 3, is constructed with this 
scale. 

Bemark. — If the scale. Fig. 2, is used as a reduc- 
tion scale in which O A represents 1 ft., we shall 
have to divide O A into 12 equal parts (inches), etc., 
and the scale will represent ^i of actual dimension. 



Fig. 1. 



Sivcft ^ivlieii 'in,64''-' 



i^ 



7 6 8 * 3 2 i 



Pig- 2. Sl/ebu'cWfl'n Soate'un.i&ei;, i"2ob 



^ 



Fig. 2. A 



'SJI.ii. A 



I'late 14. 



(b^u 



A B = 


5. Ft 


BB 


_ 


3 6 S.pt 


A C = 


27 ■ 


CC 





51. » 


A D = 


4 8.5- 


DD' 


— 


7 5 •' 


A E = 


5 8. ■■ 


EE" 


_ 


25 5 - 


A F = 


1 5.5- 


FF' 





3 2 


AG = 


10 9.- 


GG' 


_ 


6 3. " 


A H = 


1 1 7. ■ 


HH' 





5 0.5 " 


A 1 = 


i 2 8. - 


1 r 


_ 


32 " 


AJ = 


13 9. ■■ 


iJ J' 





5 0.5 " 


A K = 


152" 


KK' 





31.5 " 


A L = 


16 3." 


LL" 


_ 


10 9.5 ' 


AM = 


22 1." 


MM' 


= 


9 ■ 



H' J' 



, Fig- 3- 




Hanstein's Constructive Drawing. 



DIVISION OP CIRCLES. 

67.— Fig. 3.— Problem.— To inscribe a regular triangle, hexagon 
and dodecagon in a given circle. 

Solution.— het A B F D be the given circle. Describe with 
point A as a center and radius A C the arc BCD and draw line 
B D, which is the side o£ the required regular inscribed triangle. 

Hexagon.— Line B A — radius, B C — the side of the required 
regular inscribed hexagon. 

DodecOffon.— Bisect the arc B A by point E; draw B B, which 
is the side of the required regular inscribed dodecagon. 

68.— Fig. 3.— Problem.— To inscribe in a given circle, C, a 
square, octagon and a regular polygon of IS sides . 

Solution. — Construct two perpendicular diameters, A B and 
D G. Draw D B, which is the side of the inscribed square. 

Octago7i. — Bisect the quadrant D A (in E) and draw D E, which 
is the side of the required regular inscribed octagon. 

Tlie regular polygon of IB sides.— Bisect the arc D E by F, and 
draw D F, which is the side of the required regular inscribed 
polygon of 18 sides. 

59.— Fig. 4.— Problem.— To inscribe a regular pentai)on and 
decagon in a given circle. 

Solution.— Draw two perpendicular diameters, A B and E I, 
in the given circle C. Bisect radius C B at point D, and with 
D E as radius, D as center, describe arc E F and draw line B G 
= E F, which is the side of the required regular inscribed pen- 
tagon. 

Decagon. — Bisect the arc E G by point H and draw E H, which 
is the side of the required regular inscribed decagon. 



60.— Fig. 5.— Problem.— To inscribe a regular heptagon and a 
regular polygon of 14, sides in a given circle. 

Solution. — Draw a radius, A C. With point A as center and 
A C as radius describe arc B C D and draw BBD. HF = FD — 
D E = the side of the regular heptagon in the given circle. 

Bisect arc F H by point Q and draw H G, which is the side of 
the regular polygon of 14 sides in the circle. 



HECTIFICATION OF ARCS. 

61.— Fig 6.— Problem.— To rectify a given arc. 

Solution.— Let A B, corresponding to angle A O B, be the 
given arc. Bisect angle A O B by O N and bisect also angle 
A O N by O N'. Erect B D perpendicular to O B at B, D' D per- 
pendicular to O N at D, D' G perpendicular to O N' at D', and 
draw arc D' H with radius O D' and center O. Divide H G into 
three equal parts, and from the first division point J, near H, 
drop J L, a perpendicular to O B, then J L = arc B (J A. The 
approsimation is very close as long as the given angle does not 
exceed 60°; but for greater angles, the half of them may be 
rectified. 

From the rectified arc we can find the area of the correspond- 
ing sector: construct a triangle with the rectified arc J L as 
base and with the radius of the circle as the altitude; this tri- 
angle has the same area as the sector in question. — To transform 
a circle into an equivalent square, we may rectify the arc of 
45°, construct a triangle that has for a base 8 times the length 
of this arc, and for the altitude the radius. Transform this 
triangle into a square, then this square will be equal to the area 
of the circle. — In order to find the length of the circumference 
of a circle we would rectify the arc of 45° and multiply this 
length by 8 



Fig- 1. 





Plate 15. 
Fig. 3. 





Pi.?- 6 





Hanslein's Constructive Draiving. 



62. — Fig. 1.— Problem. — To construct a line equal to 
the semi-circumference of a given circle. 

Solution. — In the given circle C draw two perpen- 
dicular diameters, A B and F G, and, at G, the in- 
definite line E H perpendicular to F G. With A as 
center and A C as radius describe arc C D and draw 
line C D E. Make E 3 = 3 A G and draw F 3 = G 
H, which is equal to the semi-circumference of the 
circle O. Calculation gives — 

F 3 = 3.U153 times radius; 
error = 0.00006 of semi-circumference. 
Denoting the ratio of the circumference to the 
diameter of a circle by the letter 77, then this ratio 
has been more accurately found to be 

77 = 3.1415926; 
for common usage it suffices to take for it — 
77= V =3.1428, with an error = 0.001. 
Among the many approximative methods to rec- 
tify a circle, the above method has the advantage 
that it can be performed with one opening of the 
compasses. 

TANGENTS. 

63.— Fig. 2.— Problem — To construct a tangent at a 
given point of a circle. 

Definition. — A tangent is a line touching the cir- 
cumference of a circle in one point only, the point 
of contact, and is a perpendicular to a radius, 
drawn to the point of contact. 

Solution. — Let C be the given circle and A the 
point of contact. Draw the radius C A, and per- 
pendicular to it, at point A, the line M N, which is 
the required tangent. 



64. — Fig. 3. — Problem. — From a given point outside 
a circle to draw tangents to this circle. 

Solution. — Let C be the given circle and A the 
outside point. Draw A C, and on A C as a dia- 
meter describe a circle, center B ; this circle B in- 
tersects circle C at points O and P; then lines A O 
and A P are tangents to circle B. 

65.— Pig. 4. — Problem, — To construct common ex- 
terior tangents to two given circles. 
Solution. — Let C and A be the given circles. Draw 
line C A and upon this as diameter, circle C A; 
with the difference C F, of the radii D A and C E 
draw H F I from center C, intersecting circle B at 
points H and I. Draw radius C O through H, and 
C P through I. Radii A O' and A P' are parallel 
to C O and C P respectively. O' O and P' P ar& 
the points of contact of the common tangents. 

66. — Fig. 5. — Problem. — To construct common interior 
tangents to two circles. 

Solution. — Follow the previous construction and 
describe the circle C F A G. With the sum of the 
radii of both circles AD + CI = CE draw arc 
F E G; also the lines C F and its parallel radius 
A P', and C G and its parallel A O'. The inter- 
sections O and O', P and P' are the points of con- 
tact of the required tangents P P' and O O'. 




Plate 16. 
Fig. 2. 




Fig. 3. 




Fig. 4. 





HansUivi's Constructive Drauring. 



TANGENTIAL CIRCLES. 

67.— Fig. 1.— Problem — To construct circles, D and 
H, that touch a given line, M N, and a given circle 
in point A. 

Solution. — Draw line H C A B through center C 
and the given point of contact A; at A erect a per- 
pendicular to H B, intersecting M N in point E. 
With E as center, E A as radius, draw the semi- 
circle FAG and erect at F and G perpendiculars 
to M N, to obtain on line H B the intersections H 
and D, which are the centers, and H A and D A the 
radii respectively of the required tangential circles. 

68.— Fig. 2.— Problem.— To construct a circle of a 
given radius that touches a given circle and a given 
line. 

Solution. — Let C be the given circle. M N the 
given line, and R S the given radius of the re- 
quired circle. 

Draw with distance R S the line R' O parallel to 
M N. With C A = C B-f R S = R B as radius and 
center C, cut line R' O in point A, which is the cen- 
ter, and A B = R' S, the radius of the tangential 
circle. 

69. — Fig. 3. — Problem. — Within a given triangle to 
inscribe a circle. 

Solution. — Let A B C be the given triangle. Bisect 
two angles, A and C, by A D and C D, which inter- 
sect in D. Draw the perpendicular D E, which is 
the radius, and D is the center for the inscribed 
circle. 



70. — Fig. 4.— Problem. — To circumscribe about a given 
triangle a circle. 

Solution. — Let B A D be the given triangle. Bisect 
two of the sides by perpendiculars, which intersect 
in the center of the required circle. 

71. — Fig. 5. — Problem. — To connect any number of 
points by a regular curve. 

Solution. — Let ABODE, etc., be the given points. 
Draw lines A B, B C, C D, etc., and bisect each by 
a perpendicular. Take an arbitrary point G at the 
bisection line G N as a center, and with G A as a 
radius draw the arc A B; draw then B G H, a line 
to intersect the bisecting perpendicular of B C in 
H, the center, and H B the radius of the arc B C; 
I is the center, radius I for arc C D, etc. Com- 
plete the required curve to point F. 

72. — Fig. 6.— Problem —To construct a curve to tlie 
base of an Ionic column. 

Solution. — Let A D and D H be the given di- 
mensions. Trisect A D and draw in B (1st 3d) 
a perpendicular, K B E; B A is the radius 
and B the center of quadrant A K. Make B E, and 
E F = B N =i B A and draiv F E N L; E is the 
center, E K the radius for arc K L. Erect at H a 
perpendicular, H G, indefinite, at which make 
H I = L F, and draw and bisect F I by the perpen- 
dicular M J, which produced will give the intersec- 
tion point G; draw line G F O. With F as center, 
F L as radius, describe arc L O; with G as center, 
G O as radius, the arc O H. 





Plate 17. 
Fl^. 3. 







Haii:;t€in's Constructive Draining. 



73.— Fig. 1.— Problem. — To construct three tangential 
circles when their radii are given. 
Solution. — Let A, B and C be the given radii. 
Draw line G F E ^ A + B. Describe circle G with 
radius G F = A, and circle E with radius E F = B. 
With G as center, and A + C as radius, E as cen- 
ter, B + C as radius, draw arcs intersecting at H. 
H I is the radius and H the center for the third re- 
quired tangential circle. 

74.— Fig. 2.— Problem. — To construct three tangential 
circles when the three centers are given. 
Solution. — Let A B C be the given centers. Con- 
struct the triangle ABC. Make C D = C B, A E 
:= A B, and bisect D E in point F. Describe the 
required circles from points C, A and B, as centers, 
with radii C F, A F and B H. 

75. — Figs. 3 and 4.— Problem — To construct tangen- 
tial circles within a given angle. 
Solution. — Let A B C be the given angle, which is 
bisected by A D. Draw a perpendicular line D C 
at an arbitrary point D to form angle D C A, which 
is bisected by C E. The intersection of A D and 
C E is point E; with E as center, and with the radius 
E D describe the tangential circle D F. Perpen- 
dicular to A D, at point F, draw P G, and parallel 
with C E, G H. H is the center, H F the radius for 
the next circle, etc., etc. 



76.— Pig. i.— Solution 2.— Bisect the angle BAG 
by A D and at an arbitrary point, E, erect the 
perpendicular E P. Make F H = E P and draw 
perpendicular to A B at H, H I. I is the cen- 
ter, I E the radius of the circle E H J. Re- 
peat this construction by making L M ^ L J, 
etc., etc. 

77.— Figs. 5 and G.^Problem. — To construct any 
number of equal tangential circles within a given 
circle. 

Solution. — Let C A B D be the given circle. Divide 
the circle into double the number of equal parts as 
you intend to draw circles therein; for 3 circles into 
C, for 5 circles into 10 equal parts. 

Construct at an intersection of diameter and cir- 
cumference point A a tangent to intersect the pro- 
duced adjoining diameter in E. Bisect angle AEG 
by E F ; F is the center, P A the radius for one re- 
quired circle. With center G of the given circle 
and radius G F draw circle P I H, to obtain I and 
H, the centers of the required remaining tangential 
circles. 

Problem Pig. 6 is solved in a similar manner. 




Fiff- 2. 




Plate 18. 
Fig. 3, 




Fig. 4. 



Fig. 5. 






Hanstein's Constnictive Di-nwiutj. 



TANGENTIAL CIRCLES. 

78. — Fig. 1. — Problt:!!. — To divide the surface of a 
circle into three equivalent parts bounded by sejni- 
circles. 
Solution. — Let C be the given circle. Divide the 

diameter D A into 6 equal parts, and describe with 

1 and 5 as centers, 1 D as radius, the semicircles 

2 D and 4 A, with 2 and 4 as centers, and 2 D as 
radius, the semicircles D 4 and 3 A; D4A2 = iof 
area of circle. 

79.— Fig. 2.— ProbSem.— Tb construct a rosette of four 
units within a given circle. 

Solution. — Let A B be the diameter of the given 
circle. Draw four equal tangential oii'cles within 
the given circle (See Figs. 5 and 6, Plate IS) and 
connect their centers by the lines F E, E D, D H, 
and H F, which at I, I', I", and I'" pass through 
their points of contact. 

Concentric arcs may beaddedto indicate material . 
80.— Fig. 3.— ProbSem. — To construct three tangential 
circles within a semicircle. 

Solution. —Let A D B be the given semi-circle. 
Divide the radius C D into 4 equal parts, erect at 
point 1, E F perpendicular to C D, and describe 
with C as center, and radius C 3, the arc E 3 F. 
Point 3 is center, 3 D the radius to circle C D, and 
E and P are the centers to the required tangential 
remaining circles. 

A and B are the centers, A B the radius to arcs 
A G and G B, which form a Gothic arch. 
81 — Fig. 4. — Problem. — To construct two semicircles 
andthreecirclestangential within a givensemicircle. 

Solution. — Let A 3 B be the given semicircle. 
Divide radius C 3 into 3, the diameter A B into 4 



equal parts; erect at E and F, E H and F G perpen- 
dicular to A B, and at 3, H G perpendicular to C 3. 
E F are the centers, E A the radius to semicircles 
A C and C B; 3 and 4 centers, 2 — 3 the radius to 
circles 3 — i and 4 — 3, and H and G the centers for 
the required remaining tangential circles. 

GOTHIC AND PERSIAN ARCHES. 

83.— Fig. 5.— Problem -To construct a Gothic arch 
on an equilateral triangle. [Inscribe a tangential 
circle. ) 

Solution. — Let A C B be the equilateral triangle. 
Describe with B and A as centers, and radius A B 
the arcs A C and CB; AECHB is the required 
Gothic arch. 

Center F of a tangential circle in this arch is 
found by making D G=BA, DE = BG, and draw- 
ing E B, intersecting C G, in F; the center P and 
radius F G give the required tangential circle. 

Bemark. ^Whexi A R represents the thicliness of 
the stone required in work, the arcs R S and S T 
are concentric with A C and C B. The lines repre- 
senting the joints of stones, as N B ( voussoir- 
lines ), are radii in the corresponding sector. 

83. — Fjg. 6. — Problem. -To construct a Gothic arch 
w'len span and altitude are given. 
Solution. — Let A B be the given span and D E the 
altitude. Construct an isosceles triangle, A E B, 
with A B as base and D E as altitude; bisect A E 
by the perpendicular L I, whiich intersects span A B 
in I. I and K are the centers, I A the radius to arcs 
A E and E B. Make D F = A I, and P G = D I, 
and draw G I, intersecting D E, in H, the center, 
H D, the radius to the tangential circle in arch 
A E E. 





Plate 19. 

Fig. 3. 




Fig. 5. 






HanaUin' i Camilrur.tief Drauilnff. 



84. — Fig. 1. — Solution 2. — Let A B be the span and 
C D the given altitude. Construct an isosceles 
triangle, A D B, in which the base = A B, the 
altitude = C D. Bisect A D by the perpendic- 
ular L I, intersecting- the produced span in I; 
I and J are the centers, I A is the radius to 
arcs A D and B D. A D B is the required Gothic 
arch. To find center H for the inscribed circle, 
make CE = AI, EF = CI and draw F H I. 

85. — Fig. 2. — Problem To constnict a Gothic arch 

{wood or stone) with application of previous con- 
structions for its inside ornamentation. 
Remark. — This problem is intended as a review 
of former constructions, and should be drawn not 
less than three times the size of Fig. 2, to avoid in- 
accurate work by crowded lines. 

86.— Fig. 3. — Problem. — To construct a Persian arch 

about an equilateral triangle. 
Solution. — Let A D B be the equilateral triangle. 
Divide A D into 3 equal parts and draw through 
point 2, parallel with D B, G 2 E, intersecting GH 
in G and A B in E. Make D H = D G, and draw 
H F parallel to D A. E and F are centers to arcs 
A 3 and B I, and G and H the centers to arcs 2 D 
and ID; A2DIBis the required Persian arch. 
87.— Fig. 4. — Problem. — To construct a Persian arch 

whenAlB, the span, and CD, the altitude, are 

given. 



Solution. — Construct with span A B as base, and 
with altitude C D the isosceles triangle A D B. 
Trisect A D and erect in point 1 the perpendicular 
1 E; draw E 3 G, intersecting G H (parallel to A B) 
in G. Continue as in the previons construction and 
obtain the required Persian arch. 

egg-lines. 
88.— Pig. 5. — Problem. — To construct an egg-line on 
a given circle. 
Solution. — Let C be the given circle. Draw per- 
pendicular diameters A B and C D, also lines B D 
E and A D F ; B and A are centers, radius = A B 
to arcs A E and B F, and D center to arc E F ; 
A B P E is the required egg-line. To obtain a more 
elongated shape of an egg-line, place centers A' B' 
further out, but equidistant from C, and describe 
arcs A E' and B P', and with D as center arc E' F'. 

89.— Fig. 6. — Problem. — To constmct an egg-line 
when the short axis is given. 

Bemark. — The longest line possible to be drawn 
in the egg-line is called its long axis, and the 
greatest width perpendicular to it is the short axis. 

Solution. — Bisect the given short axis A B by 
the perpendicular D E, on which make H C J. 
C 1 1 of A B; C P = C G = -g of A B; F and G are 
centers, and P B the radius to arcs L N and K J, 
H to K L and ItoJN;KLNJis the required egg- 
line. 



Fig. 1. 




Fis- 2. 




Plate 20. 

Pig. 3. 







i1aii^t€ln'3 C^nslmclivc Drawing. 



90.— Fig. 1.— Problem,— To construct an oval or lens-Une at ad- 
joining equal squares. 

Definition.— An oval is an elongated endless curve consisting 
of symmetric arcs. The longest possible line drawn in an oval 
is called its long o.cis, and the greatest width perpendicular to 
it is called its short axis. Both axes divide the oval into sym- 
metric parts. 

Solution.— Let A F G C and F G D B be the given squares. 
Draw the diagonals F C and A G, intersecting in H, and F D 
and B C, intersecting in I; G and F are centers, G A, the radius 
to arcs A B and O D, H and I the centers to arcs A C and B D. 

91.— Fio. 2.— Problem.— To construct an oval at a given circle. 

Solution.— liSt A B F G be the given circle. Construct two 
perpendicular diameters, A F and B G, and draw A B D, A G I, 
FEE and F G H; F and A are the centers, radius F A to arcs 
E A H and D F I; B and G are the centers, radius B D to arcs 
E D and HI; E H I D is the required OTal. 

93.- Fig. 3.— ProbEsm.— To co7istnict an oval, at two equal 

circles, of which the circumference of one passes through the 

center of the other. 

Solution. — Let A and be the given circles, intersecting each 

Other in B and D. Draw from points B and D through centers 

A and C, lines BAG, BCH, DAF and DOB. D and B are 

the centers, radius D F to arcs F E, and G H. F B H G is the 

required oval. 

83.— Fig. 4.— Problem.— To construct an oval when its long and 

short axes are given. 

Solution. — Let A B and C D, bisecting perpendicularly, be the 

long and short axis respectively. Draw C B and the quadrant 

C K from center E. Make C N = K B and bisect N B by the 



perpendicular O L H; I E = B H, and E J = L B, and draw 
H J P, IJ E and I L S. J and L are the centers, J A the radius 
to arcs P E and O S, and H and I the centers to area P O and 
li S; P O S R is the required oval. 

ARCHES. 

94.— Figs. Band 6. —Problem.— To consfT-uct an arch, its span 
and altitude being given. 
Solution.— het A B be the given span, and C D, the perpen- 
dicular in its bisection point C, the altitude. Construct with H 
A B = A C and C D the rectangle C D E A, and draw diagonal 
D A. Bisect angles EDA and E A D by F D and F A. From 
F, perpendicular to A D, draw F H Q and make I C = H C; 
H and I are the centers, with radius H A to arcs A F and 
B J, and G the center, radius G F to arcs FDJ; AFDJBis 
the required arch. 

Remar/c.- When we assume the thickness of the stone used 
in the arch as B O, we describe the concentric arcs O N, N E 
and E L, and divide these into equal parts, except keystone K, 
to which generally more prominence is given. As in the Gothic 
arches, the joint lines of the stones are radii in the correspond- 
ing sector. 

95 —Fig. 6.— Solution 2.— Let A B be the given span, and D 
the altitude. With E A as a radius shorter than the given 
altitude, and centers E, D and F, describe the circles E A, 
D G and F B ; draw and bisect E G, intersecting the pro- 
duced altitude in point H, the center, with radius H I to arc 
I D L. Complete the required arch A 1 D L B and add its 
stone units. 



Plate 21. 



Fig. 1 




Fig. 2. 





Fig. 4. 



Fig. 5. 





Fig 6. 




UanstcLn'n Cmistnictivo Drawirnf. 



96.— Fig. 1. — Solution 3. — Let A B be the span, and 
C D the altitude. Construct with A C the equi- 
lateral triangle A E C, and make C F = C D, 
and draw D F G. Parallel with E C draw G H I; 
points H and K are the centers, A H the radius 
to arcs A G and J B, and I the center, I G the 
radius to arc G D J. Proceed as in Fig. 6, and 
complete the required arch and its stone units. 
97. — Pig. 2. — Solution 4. — Let A B be the span, and 
C D the altitude. Construct on altitude C D 
the equilateral triangle DEC, make C F=C A, 
etc., and proceed and complete as in Pig. 1. 
98.— Pig. 4.— Problem. — To construct an elliptic arch 
when span and altitude are given. 
Solution. — Let A B be the given base, C D the 
altitude. Produce A B, and with radius C D' = 
C D 1= the given altitude describe semicircle JD' A. 
Divide J A and span A B into the same number of 
equal parts, and erect at all division points per- 
pendiculars. With the T square make C D = C D', 
S E' and 4 E" = 2 E, 1 G' and 5 G" = 1 G, etc., and 



connect points B G" E" D E' G' A by a free-hand 
line, and complete the required elliptic arch. 
99. — Figs. 3 and 5.— Problem. — To construct ascend- 
ing arches when span and altitude are given. 
Solution. — Let A B be the given span and C B the 
altitude; draw CA, the ascending line. Make BD 
(the produced span) = B C, and bisect A D by the 
perpendicular E G; E is center, E A the radius to 
quadrant A G, and P the center, P G the radius to 
quandrant G C. A G C B is the required arch. 
Complete and add the stone units as in previous 
constructions. 

100.— Fig . 5.—Solution2. — Let AB be the given span, 
and B D the altitude. Draw D A, the ascending 
line, and bisect A B by the perpendicular P C; 
bisect angle F E A by G J, and with J as cen- 
ter, J A as radius, describe arc A F. Draw P J 
and D H parallel to A B, and with center H, 
radius H D describe arc P D; A P D B is the 
required ascending arch. Complete and add 
stone units as in previous constructions. 



Plate 22. 





i.-ii 







Fig.- 4. 




Fig. 5. 



Av'^'l "v""' 





^'A—/.! -4^^' \ 




1 7,'^^~~i~~~^i 





Hanstein'g Constructive Drawing. 



IONIC SPIRALS. 

101.— Figs. 1 and 1 A.— Problem^— To construct an 
Ionic spiral luhen the altitude is given. 

Solution. — Let A B be the given altitude. Divide 
A B into 16 equal parts. The center of the spiral 
eye is situated in the 9th part from B, and its 
radius = j^^ of A B. 

Pig. 1 A. — Bemark. — To explain division and sub- 
division, the eye of the spiral in double size is 
represented in Fig. 1 A. It is advisable to exe- 
cute Fig. 1, Plate 23 and Fig. 1, Plate 34 in as 
large a scale as possible, to facilitate an ac- 
curate division and subdivision. 

Draw two perpendicular diameters, D G and 
FH, and inscribe the square DF GH; inscribe 
in this the square 1, 2, 3 and 4. Divide the dia- 
gonals 1 3 and 3 4 into 6 equal parts and draw 
the squares 5, 6, 7 and 8, and 9, 10, 11 and 12. 

The center of the first quadrant is point 12, 
the radius 12 D, describe D I; 11 the center, 11 
I the radius to quadrant I K, etc., and go back 
as the numbers indicate, 10, 9, 8, 7, 6, 5, 4, 3, 3, 
until the center of the last quadrant is point 1, 
the radius 1 J to quadrant J B. To obtain the 
second curve, we trisect the distances 12 C, 11 C, 
IOC, 90; also 5, 9 — 6, 10 — 7, 11, etc., etc., and 
with the center in the first 3d from 13 towards 
C, draw the first quadrant ; first 8d, from 11 
towards C, as center the next quadrant, etc., 
etc., and complete the curve in the same order, 
locating the centers at diagonals in the first 3d 
from the original division towards C. 



ELLIPTIC ASCENDING ARCHES. 

102.— Fig. 2.— Problem.— To construct an elliptic aa- 
cending arch when span and altitude are given. 

Solution. — Let A B be the span and B C the alti- 
tude. Draw C A, the ascending line, and describe 
on A B as diameter, a semicircle, A NO B. Divide 
the diameter into any number of equal parts (6) and 
erect in each division perpendiculars, at which we 
make2'N'=2N, 4'P = 4PandN' R' = N R and 
connect R' O' P' N', etc., by a free-hand line, which 
is the required arch. 

Bemarh. — This curve is also applied at the base 
of the Ionic column, as Fig. 6, Plate 9. 

103.— Fig. 3.— Problem.— To construct an elliptic as- 
cending arch when span, its ascending and mean 
altitudes are given. 
Solution. — Let A B be the given span, B C the 
ascending and E F the mean altitude. With the 
mean altitude E F ^ E' F' describe the quadrant 
F' H A E'; divide radius E' A in 3 and subdivide 
the last 3d into 3 equal parts. Divide the span into 
the same number of proportional parts and erect 
perpendiculars. Transfer the altitudes of F' H J, 
etc., to the perpendicular A D, and draw lines 
parallel with the ascending line A 0, to obtain the 
points of intersection J' J", H' H", F, etc., which 
points, connected by a free-hand line, will give the 
required arch, C J" H" F H' J' A. 




Fig. 1. A 



Plate 23 . 
Fig. 2. 



/l M ^12- 


^ 1* 1 1 ^ 

1 !ir 


* 1 2 


1 5 NB 



=-,;§■. 3. 




HiimUin's Constmclll'C Oici 



104.— Figs. 1 and lA. — Solution.— Let A B be the 
given altitude of the spiral, which is divided 
into 14 equal parts. Point C, the center of the 
eye of the spiral, is located at the 8th part 
from B; its radius ^ of A B. Construct the 
square D F E H (see Fig. 8), and inscribe the 
square 1, 2, 3 and 4; bisect C 1, C 2, C 3 and 
C 4; bisect again C 5, C 6, C 7 and C 8 and 
draw the squares 5, 6, 7, 8 and 9, 10, 11, 13. 
Describe, first, the quadrant D I, from center 
12 and radius 13 D, 3nd, quadrant I K, from 
center 11, with radius 11 I, etc., as operated in 
Fig. 6, until point 1 is the center, I J the radius 
of the last quadrant J B of the third revolu- 
tion. The subdivision for the centers of the 
second curve is as follows: 

Bisect C 12, C U, C 10 and C 9, and bisect also 
12, 8 — 11, 7 — 10, 6 and 9, 5, and make these bisec- 
tion points vertices of squares parallel to the 
square 1, 3, 3 and 4. Divide further the lines 1, 5 
■^ 3, 6 — 8, 7 and 4, 8 into 4 equal parts, and con- 
struct a square parallel to the vertices located in 
the first 4th from points 1, 3, 3 and 4 towards C. 



The vertices of the squares of these subdivisions 
are the centers for quadrants of the second curve, 
which quadrants are described as in Fig. 6. 



105.— Fig. 3.— Problem.— To construct a spiral with 
semi-circles when the spiral " eye" is given. 

Solution. — Let C, a small circle, be the given 
spiral eye. Draw and produce a horizontal diam- 
eter, M A B N. With A as center, A B as radius, 
describe the semi-circle BO; C as center, C O as 
radius, semi-circle OP; A as center, A P as radius, 
semi-circle P R, etc. Curve B O P R, etc., is the 
required spiral. 

106.— Fig. 3.— Problem.— To construct the evolute of 
a given triangle. 

Solution. — Let A B C be the given triangle. Pro- 
duce C B, B A and AC; B is the center, B A the 
radius to arc A N; C the center, C N the radius to 
arc N O; A the center, radius A O to arc O P, etc., 
etc. Curve A N O P, etc., is the required evolute. 



Plate 24. 




Han8tein'8 Constructive Drawing. 



CAM-LINES— ARCHIMEDEAN SPIRALS. 

Definition.— An archimedean spiral is a curve in a plane 
generated by a point whose distance from a centre of rotation 
increases uniformly. 

Cams are arrangements in mechanics by which a rotary 
motion is converted into a reciprocating action, they are con- 
structed by archimedean spirals. 

Remark. — The following curves, used principally in mechan- 
ics and architecture, should be executed by free-hand hnes 
before the student attempts to use a curve rule. 

107.— Fig. 1.— Problem.— To construct a cam-line of IM revolutions 
when the distance C between revolutions is given. 
Solution.— With 8 equal parts, 6 of which are equal to the 

fiven distance C C, describe the circle 8 A B D E F, which is 
ivided into 6 equal parts by diameters. Describe circles with 
C as center, radius C i, to intersect diameter B F in B'; with 
radius C3 to intersect D 8 in D'; C 3 to intersect E AinE', etc.; 
connect points C B' D' E' F' 5 C H D by a free-hand line, to 
complete the required cam-line. 

108.— Fig. 2.— Problem.— To construct a heart-shaped cam when the 
altitude is given. 

Remark. — Heart-shaped cams are made to convert half of a 
revolution into forward motion, the other half of the revolution 
into backward motion. ( Piston-rods for pumps, etc.) 

Solution. — Let C 8 be the given altitude, which is divided into 
8 equal parts and is the radius, C the center of the circle, 
divided by diameters into 16 equal parts. With center C, radius 
C 1, describe circle to intersect radii C A and C G in A and A', 
with C 2 as radius to cut radii B C and J C in B' and B", with 
C 3 to cut radii C C and K C in C a.nd O", etc. Connect 
C A' B' C D' I E H C" B" A" C by a free-hand line and complete 
the required heart-shaped cam. 

109.— Fig. 3.— Problem.- To consti^ict a cam in 4 equal divisions, to 
ral^e a lever in the first 14 of its revolution^ equal to the altitude 
B D, to remain stationary the second }i, to descend its first 
position the third %. and remain stationary the last J4. of its 
revolution. 

Solution. — Let B D be the given altitude, B A an arbitrary 
tance from the huh, and C the center of the cam. Describe with 
C D, center C, the circle D H D' 4 and divide it into quadrants, 
two opposite ones into 4 equal parts again, by diameters. 
N 4 = B D = the given altitude is also divided into 4 equal parts, 
3, 2, 3 and 4, and with radius C 1 draw arcs 1 G', G, with radius 



C 2. 2 F' F, wit radius C 3, 3 E' E; connect N G' F' E' D' and 
the symmetric points B G F E H by a free-hand line and com- 
plete the required cam. 

110.— Fig. 4.— Problem,To construct a cam in three equal divisions^ 
which in one revolution shall lift a lever -= A4 in theffrst 3d, 
shall remain stationary the second 3d, and shall rise again the 
third 3d an altitude = 4 B and make a suddeyi escape at B, to 
renew its motion in the second revolution. 

Solution. — Let the two inner circles be shaft and hub circum- 
ferences, A 4 the altitude of the first incline, 4 B the altitude of 
the second incline (the third division). 

Remxtrk.—This construction, in applying the principles of 
Figs. 1, 2 and 3, will not present any difficulty to the student, 
and can now be solved without the assistance of a teacher. 

CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA. 



111. — Fig. 5. — Three curves, which we obtain by sectional planes 
through a circular cone and cylinder, are of the greatest im- 
portance in technical work; the ellipse, pa7'ai)olaand hyperbola. 
A sectional plane through the cylinder or circular cone in an 
oblique direction, as U V or M N, respectively, creates the 
ellipse. A sectional plane S T, parallel to the side C B of the 
circular cone, creates the parabola. A sectional plane Q R, 
parallel to the axis of the circular cone, creates the hyperbola. 
Definition.— An ellipse is a closed curve; the sum of the dis- 
tances of each point in this curve from two fised points within, 
called foci, is equal to the long axis. The ellipse has two axes, 
the major and minor, bisecting each other perpendicularly and 
dividing the ellipse as well as its surface into two symmetric 
parts. 

112.— Fig. 6.— Problem.— 2'o construct an ellipse when major axis 
(ti^ansversant) and minor axis (conjugant) are given. 

Sohttion. — Let A B be the major, C D the minor axis. Wlieu 
a radius i^ A B = A M and center C draw arc and intersections 
with A B, points F and F', the foci; divide F M arbitrarily into 
parts, increasing in length towards M. and with F and F' as 
centers, B 4 as radius, describe arcs E G and E' G'; with F and 
F' as centers, A 4 as radius, draw intersections at E and G and 
at E' and G'. Points E E' G G' are situated at the circumference 
of the ellipse, Operate with points 3. 2 and 1 in the same 
manner, and we obtain by each operation 4 points, which lie at 
the circumference of the ellipse, as with point 3, e. g,, by which 
we locate points H J H' J'. Connecting these points by a free 
hand line, we obtain C E H A J G, etc., the required ellipse. 




Fig. 2. 







Pl ate 25. 
■Fig. 3. 




Fig. 4. 





iiT 




l|k'^. 


— '' / ' 



Jffanstein's Co7\f>tructive Drawing, 



313.— Fig. 1.— Problem.— To consfract a tangent to an ellipse 
when the point of eontact is given. 

Solution.— liet A C B D be the ellipse and Q the point of con- 
tact. Describe from G as center, with.radius G F, the arc F N, 
and draw and produce line F' G, intersecting arc F N in N; bi- 
sect angle N G F by IJ, whicli is the required tangent. 

Bemarlz. — In elliptic arches, executed in cut stone, the joints 
are perpendiculars (as P G) to tangents, having the unit divisions 
as points of contact. 

lU.— Fig 1.— Problem.— From cm exterior point to constmct a 
tangent to an cllipae. 
Solution.— Let H be the given exterior point. With H as 
center, H F' as radius, describe arc F' O; with A B as radius, 
and F as center, intersect are F' O in O. Bisect arc F' O by L H, 
which is the required tangent. 

115.— iFiG. 3. — ProWem.— To construct an ellipse when both axes 
are given. (Practical sohttion.) 

Solution 1. — Let A B and C D be the given axes. Find the foci 
(112) and place in F, F' and C pins, around which tie a linen 
thread to form the triangle F C F'. Take away the pin at C and 
place the pencil point in the triangle, by stretching the thread 
gently and forming a vertex of the triangle; draw the curve, 
which will be the required ellipse. 

Solutions.- AB and CD are the given axes. Take O P, a 
straight edge or a slip ot paper, at which make A' M' = A M = 
J^ A B and A' C — C M = ^4 C D. Guide the straight edge to 
have point C follow the major axis, and M' the minor axis, then 
will point A' describe the circumference of the required ellipse. 
Locate the position of point A' during this operation by pencil 
marks, which, connected, will give the ellipse. 

Bemarfc.— Place in points C and M' pins, in point A' a pencil 
point, and let these pins slide in grooves in the place of the axes; 
we have an instrument called a trammel or ellipsograph, with 
which we are able to draw any ellipse by arranging points A' C 
and M' in the required proportions. 



116.— Fig. 3.— Problem.— To construct a7i ellipse by intersecting 
lines. 

Solution.— liet A B and C D be the given axis, and construct 
with these lines the rectangle B F G H; divide A B and E G into 
the same number of equal parts and number as in the diagram* 
Draw lines D 1 P, D 3 O and D 3 N, intersecting the lines C 1, C 2 
and O 3 at P, O and N, etc., which points, connected by a free- 
hand line, will be the required ellipse. 

117.— Fig. 4.— Problem.— To construct an elliptic curve in an 
oblique parallelogram. 

Solution.— Jjet E F G H be the parallelogram. Draw axes A B 
and C D bisecting opposite sides, and divide C M and E C into 
the same number of equal parts; proceed as in the previous 
construction and draw C P O N A, etc., the required ellipse. 

118.— Fig. B A.— Problem.— To construct an ellipse by intersec- 
tions of lines. 

Solution.— With A B and C D, the given axes, construct the 
rectangle E F H Q; divide E O and A E in the same number ot 
equal parts (4) and number as shown in the diagram. Draw 
lines 1 A, 3 3, K 2, and C 1, and connect their intersections T S R, 
etc., by a free-hand line to complete C T 8 R A, etc., the re- 
quired ellipse. 

119.— FiG.5 B.— Problem.- To construct an ellipse by its tangents. 

Solution. — Draw and divide C B into any number of equal 
parts (4): 1, 3, 3 and 4, through which parallel with C D draw 
P P', O O' and L L'; draw also E 1 1, E 2 K and ESN and lines 
L N, O K and P I, which are the tangents to the required ellipse. 
Draw the ellipse by a free-hand line. 

120.— Fig. 0.— Problem.— To cojisfruct am, ellipse by the differ- 
ences of two circles. 

Solution, — Let B A and C D be the given axes. Describe with 
B A and C D as diameters concentric circles with center M. 
Divide both circles into 13 equal parts by the diameters 10, 4 — 11, 
5—1, 7-3, 8 and 3, 9. Draw lines 7, 5—8, 4-10, 3 and 11, 1, and 
from the intersection points E F G and H the perpendiculars to 
10, 3—11, 1-7, 6 and 8, 4, which will give points N O A P B D, 
etc., at the circumference ot the required ellipse. 



Plate '26. 





Fig. 5. 



P'g- <5. 








'H — H 

^ 1/ 



/! ^.n 









\1 fV> 



Hanniciri's Constrxictivc Drawing. 



121.— Fig. 1.— Problem.— To construct a parabola when the axis 
and the base are given. 

Definition.— The parabola is a curve in which the distance of 
any point from an outside right line (directrix) is equal to the 
distance of this point from a fixed point within, called focus. A 
line bisected perpendioularly by the axis at its terminus and 
Intersecting the curve is called the base, and a parallel with it, 
through the focus, the parameter of the parabola. 

Solution.- Let AP be the axis and Lli. tlie given base. Bisect 
IjP ■= Yz the base L K in J, and draw J A. In J erect a perpen- 
dicular to J A, J R intersecting the produced axis in E; trans- 
fer PR to left and right of point A, to obtain point F, the focus, 
and point O, through which draw M N, the directrix, perpen- 
dicular to the axis O P. Divide A P into arbitrary parts, 1, 3, 3, 
4, etc., in which ereut perpendiculars, and with F as center, O 1 
as radius, cut the perpendicular 1 in B and B'; with 13 as ra- 
dius, the same center, cut the perpendicular 3 in G and C ; with 
O 3 as radius cut perpendicular 3 in D and D', etc., and connect 
the obtained points L E' B' A B E K by a free-hand line, which 
is the required parabola. 

122.— Fig. 3.— Problem,— To construct a tangent to a parabola 
lohen the point of contact is given. 
Solution.— Let L R K be the given parabola, O P the axis, M N 
the directrix, and A the point of contact. With A as center, 
A F as radius, draw arc F B and A B perpendicular to M N. Bi- 
sect arc F B by line S G, which is the required tangent. 

Problem.— To construct a tangent to a parabola from an ex- 
terior point, B. 

Solution.- With E as center, and B F as radius, draw arc F D 
and erect at D a perpendicular to M N, intersecting the para- 
bola in H, the point of contact ; or bisect arc D F by line T E, 
which is the required tangent. 



133.— Fig. 3.— Problem.— 2'o eonstrwct a parabola when two sym- 
metric tangents are given. 
Solution.— Let B E = A E be the given tangents. Divide E B 
and A E into equal parts and number as shown in the diagram. 
Draw lines 7-7, 6-6, 5-5, 4-4, etc., which are the tangents of the 
parabola. A free-hand curve tangential to these tangents is the 
required parabola. 

134.— Fig. 4.— Problem.— To construct a parabola when the axis 
and the base are given or the rectangle drawn with these 
lines. 
Solution.— Let A B 6 J be the given rectangle. Divide H 8 J 
= D 6 and B 6 into 6 equal parts, respectively ; number as in the 
diagram, and draw parallel to the axis D O lines through 1, 3, 3, 
4, 5. Draw also lines 5 D, 4 D, 3 D, 3 D and 1 D, intersecting with 
the horizontals in points I H G E F D, etc., which points, con- 
nected by a free-hand line, furnish the required parabola. 

135.— Fig. 5.— Problem.— To construct a parabola practically 
when base, O P, and axis, A B, are given. 
Solution.— Locate the focus F and the directrix M N and 
place a straight edge firmly coinciding with it. Fasten a thread 
to a pin placed in F and pass it around a pin in A to a point D 
of the set square, when its side C D coincides with axis A B. 
Remove the pin in A and hold the pencil to stretch the thread 
gently, touching C D constantly, shift the set square to the left. 
The pencil point will describe the required parabola on the 
drawing paper. 

126.— Fig. 6.— Problem.— To construct a Oothic arch by para- 
bolas. 
Solution.— Let A B be the span and F E the altitude of the 
arch. Construct the rectangle C D B A, divide C D into 8 and 
E F into 4 equal parts and number as the diagram. Draw lines 
1 A, 2 A, 3 A and parallel to span 1 1 1', J 3 J', and H 3 H'. The 
points of intersection, A I J H E H' J' I' B, connected by a free- 
hand line, complete the arch. 




Fig. 2. 




Plate 27. 

Fig. 3. 




Fig. 4. 



Fig 5. 





r'-i — "/ 
1 / / 


1/ 


^XT| 


i !./f 




X\) 


' /' 


^vii 


f' 


^^' 



fffmsUln's Constructive Drawing. 



127.— Fig. 1.— Problem.— To construct hyperbolas 
when the vertices and foci are given. 

Definition. — The hyperbolas are curves; the dif- 
ference of distances of each point to the foci is 
equal to an invariable line, the axis. 

Solution.— Place on line M N, A and B the ver- 
tices, and F and F' the foci equidistant from O. 
From F' towards M mark arbitrary divisions and 
number as in diagram. With radius B 1, center 
F, — radius Al and center F' draw intersecting 
arcs at C and C; radius B 2, center F and radius 
A 3 and center F' draw intersecting arcs at D' and 
D, etc. Connect G' E' D' C A C D F G by a free- 
hand line, to complete the required hyperbola. To 
obtain the second curve, operate symmetrically. 
128. — Fig. 2. — Problem. — To construct a tangent to a 
hyperbola when point of contact, P, is given. 

Solution. — Draw line P F, and with radius P F' 
and center P the arc F' D. Bisect F' D by the line 
T U, which is the required tangent to the hy- 
perbola. 

Remark. — The stone joints in hyperbolical arches 
are the perpendiculars to tangents at the point of 
contact. 

Problem. — From an exterior point, R, to construct a 
tangent to the hyperbola. 

Solution. — With R as center and radius R F 
draw arc F N; with F' as center and radius A B 



cut arc F N in N and bisect F N by S R, which is 
the required tangent to the hyperbola. 
129.— Fig. 3. — Problem. — To construct hyperbolas 
when axis A B is given; to find foci and draw the 
asymptotes. 

Asymptotes are right lines to which the branches 
of the hyperbolas do approach when produced, 
but do not touch. 

Solution. — Construct the square E D C G with 
C D = A B, which the axis divides into two equal 
rectangles. Draw and produce the diagonals M N 
and O P, which are the required asymptotes. With 
O as center, O G as radius, draw arcs G F' and 
C F. With F and F', the required foci, draw the 
hyperbolas, as in Fig. 7. 

EVOLUTE. 

130.— Fig. 4.— Problem.- To construct an evolute at 
a given circle. 

Definition.— An evolute is a curve made by the 
end of a string unwinding from a cylinder. 

Solution. — Let C be the given circle ( the section 
of a cylinder). Divide the circumference into a 
number of equal parts (12) and draw the diameters 
and tangents 1 A', 2 A", 3 A'", 4 A^ etc. With 
center 1 and radius 1 A describe arc A A'; center 
2, radius 2 A', the arc A' A"; center 3, radius 
3 A", the arc A" A'", etc.; curve A, A', A", A'" is 
the required evolute. 




Fig. 1. 



A 



Pl ate 28. 
Fig. 2. 




Fig. 3. 




Fig 4. 




Hanstein's Cdnstnictive Drawing. 



GEAR LINES— CYCLOID. 

131.— Fig. 1.— Problem — To construct a cycloid 
when the generating point A is given at the cir- 
cumference of the circle. 

Definition. — A cycloid is a curve generated by a 
point at the circumference of a circle, making one 
revolution in rolling on a straight line. The curve 
generated, when the circle rolls on the outside cir- 
cumference of another circle, is the eincycloid, and 
when the circle rolls on the inside circumference 
of another circle, the hypocycloid. 

Solution 1. — Let C be the rolling circle, tangent 
A B its rectified circumference and A the generat- 
ing point. Divide the circle C and line A B into 
the same number of equal parts (13) and number as 
in diagram. Pass horizontals through points 1, 2, 
3, etc., of the rolling circle and erect perpendicu- 
lars at A B in points 1, 3, 3, etc. With points C, 
C", C, C* as centers, C A as a radius, describe 
circles 1 A', 2 A", 3 A'", 4 A*", etc., which points 
connected give the required cycloid. 

132.— Fig. 2.— Solution 2.— Follow the operations 
of the previous construction. Draw the circle 



C« 6, also chords 6 I, 6 H, 6 G, 6 F, 6 E and 

their symmetric chords. Parallel to 6 E draw 
E' 7 K, to 6 F, F' 8 L', to 6 G, G' 9 M', to 6 H, 
H' 10 N' and to 6 I, I 11 O'. O' is the center, 
O' B the radius to arc B I'; N' the center, N' H' 
the radius to H' G'; M' the center to G' P', L 
the center to F' E' and K to E' D E, etc. Com- 
plete the construction symmetrically to the 
left of axis D K. The curve of B I' H' G', etc., 
is the required cycloid. 
When a cycloidal arch is executed in stone, the 
radii of the pertaining arcs are the joints of the 
units. 

138. — Pig. 3. — Problem.— To construct a cycloid 

when the point generating the curve is situated at a 

greater radiVjS than the rolling circle. 

Solution. — Let C G be the rolling circle, G 13 its 

rectified circumference and A the generating point. 

Describe with C A from C a concentric circle and 

proceed in this construction as in Fig. 1. Pass 

horizontals through the divisions of the greater 

circle and describe with radius A and centers 

C, C=, C^ etc., the circles B A', A", D A'", etc. 

Points A, A', A", A'", etc., connected by a curve, 

are the required cycloid. 



Plate 29. 




HansteWs Constructive Drawing. 



GEAR LINES— EPICYCLOID AND HYPOCYCLOID. 

134.^ — Fig. 1. — Problem. — To construct an epicycloid 

when the relation of the rolling to the stationary 

circle, is 1 : 2. 

Solution.— Liet A B and 6 A be the diameters of 

the given circles, having the proportion of 2 : 1, 

respectively. Divide the rolling circle into any 

number of equal parts (12), and as circumferences 

are proportional to diameters, the circumference 

of 6 A = the semi-circumference B A contains 13 

of the same equal parts. With center C draw 

circles passing through points 1, 2, 3, 4, 5, 6 and 

D, and also the diameters D'a, D"b, D"'c, etc. 

D', D", D'" are the centers and radius D A to arcs 

a A', b A", c A'", etc. 

Connect A, A', A", A'", etc., by a curve, which 
is the required epicycloid. 

185.— Fig. 2. — Problem. — To construct a hypocycloid 
wlien the relation of the circles is as 1 : 2. 
Solution.— Trea^ting this construction as the pre- 
vious one, we shall obtain a right line A B as the 
required hypocycloid. 



This construction is the fundamental principle 
of the planet wheel, applied to convert directly a 
rotation into a reciprocating movement (pump- 
piston ). 

136.— Fig. 3.— Problem — To construct a hypocycloid. 

Solution. — Let C be the circle, point A the gener- 
ating point rolling in circle E. Relation of circles 
1 : 3. Make an equal division in both circles (A b 
= A 1) and draw radii A C, b C, c C", etc. C, C, 
C", C", etc., are the centers and C A the radius to 
arcs b A', c A", d A'", etc. Connect A, A', A", 
A'" by a curve, which is the required hypocycloid. 



137. — Pig. 1. — Probiem. — To construct an epitrochoid. 

It will not be difficult to execute this curve. See 
Fig. 2, Plate 29. 

138. — Pig. 2.— Problem.— To construct a hypotro- 
choid. 

Solution.— Lieb C P be the rolling- circle, A the 
generating point and D J B the circumference on 
which circle C rolls. Proceeding as in Pigs. 2 and 
3, we obtain the curve A, A', A", A'", etc., which 
is the required hypotrochoid. 



APPLICATIONS TO ARCHITECTUBE. 

139.— Fig. 1. — Problem.— To construct a design for 
an ornamented Gothic arch in stone. 

This construction is based on principles ex- 
plained and described in the previous part of this 
volume, and its solution should not present any 
serious obstruction to the student. 

Hemark. — To obtain an accurate result, it is ad- 
visable to make the equilateral triangle, the fun- 
damental figure of this arch, not less than 8 
inches a side. 



APPLICATIONS TO MECHANICS. 

140. — Fig. 1. — Problem. — To construct a pair of 
spur-wheels, their relation tohel : 2. 

To solve this problem we require the construc- 
tion of two epicycloids and two hypocycloids to 
the '^flanks" of the teeth, and it is advisable to 
enlist the advice of a teacher, to execute this im- 
portant construction correctly. 



Hanstein's Skeleton Models and Goniostat 




A teachers claas'Toom device for practical demonstra- 
tions in Drawing, Arithmetici Plane and Solid Geometry, 
Projection, Perspective, Shadows, Stereometry, Stereotomy, 
Axonoraetry, Crystallography, and Astronomy. 

An arrangement with which the regular, and an endless jiumber of 
irregular surfaces and solids, their intersection and penetrations in 
skeleton form, may be built by the student or teacher in the same time 
it will take to make a blackboard sketch. 

The models are two feet high, easily visible by every pupil of a large 
class-room, and with the application of the "Goniostat" may be placed 
in any position in space and rotated in horizontal circles, in vertical 
circles and in any inclined plane in space. 

Address Prof. Herman Hanstein. Supervisor of Drawing Chicago 
High Schools, Director of Drawing Dep't Chicago Mechanics Institute. 

361 Mohawk Street, Chicago. 




Hanstein's 
Blackboard 

Compasses 



Warranted 
not to slip 
on any 
Blackboard 
Surface 



X. 



Manufactured and for sale by 

The Keuffel & Esser Co- 

New York Chicago St, Louis San Francisco 



PRINTERS, CHICAGO 



SEP S 1904 



